Problem: Balance the following chemical equation: $ $ $\text{CH}_4 +$ $\text{O}_2 \rightarrow$ $\text{CO}_2 +$ $\text{H}_2\text{O}$
Explanation: $\text{C}$ is already balanced. There are $4 \text{ H}$ on the left and $2$ on the right, so multiply $\text{H}_2\text{O}$ by ${2}$ $ \text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + {2}\text{H}_2\text{O} $ That gives us $4 \text{ O}$ on the right and only $2$ on the left, so multiply $\text{O}_2$ by ${2}$ . (Since oxygen is by itself on the left, it should be done at the end because you can give it a coefficient without affecting another element.) $ \text{CH}_4 + {2}\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} $ The balanced equation is: $ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} $